Find the eccentricity of the ellipse of minor axis is 2b, if the line segment joining the foci subtends an angle `2alpha` at the upper vertex. Also, find the equation of the ellipse if the major axis is 2√2

Answer

To solve the problem, we need to find the eccentricity of the ellipse given that the minor axis is \(2b\) and the line segment joining the foci subtends an angle \(2\alpha\) at the upper vertex. We also need to find the equation of the ellipse given that the major axis is \(2\sqrt{2}\).

### Step 1: Understand the properties of the ellipse
The standard form of an ellipse with a horizontal major axis is given by:
\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]
where \(2a\) is the length of the major axis and \(2b\) is the length of the minor axis.

### Step 2: Identify the values of \(a\) and \(b\)
From the problem, we know:
- The minor axis is \(2b\), so \(b = 1\) (since \(2b = 2\)).
- The major axis is \(2\sqrt{2}\), so \(a = \sqrt{2}\) (since \(2a = 2\sqrt{2}\)).

### Step 3: Use the relationship between \(a\), \(b\), and eccentricity \(e\)
The relationship between \(a\), \(b\), and eccentricity \(e\) for an ellipse is given by:
\[ b^2 = a^2(1 - e^2) \]
Substituting the values of \(a\) and \(b\):
\[ 1^2 = (\sqrt{2})^2(1 - e^2) \]
This simplifies to:
\[ 1 = 2(1 - e^2) \]
\[ 1 = 2 - 2e^2 \]
\[ 2e^2 = 2 - 1 \]
\[ 2e^2 = 1 \]
\[ e^2 = \frac{1}{2} \]
Taking the square root gives:
\[ e = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \]

### Step 4: Find the equation of the ellipse
Now that we have \(a\) and \(b\), we can write the equation of the ellipse:
\[ \frac{x^2}{(\sqrt{2})^2} + \frac{y^2}{1^2} = 1 \]
This simplifies to:
\[ \frac{x^2}{2} + y^2 = 1 \]

### Final Results
1. The eccentricity \(e\) of the ellipse is \(\frac{\sqrt{2}}{2}\).
2. The equation of the ellipse is \(\frac{x^2}{2} + y^2 = 1\).